Question

Find the zeros in simplest radical form:
   y=1/2x^2-4

Answer 1

`y= (1)/(2) x^2-4\\\\y=0\ \ \ \Leftrightarrow\ \ \ (1)/(2) x^2-4=0\ /\cdot2\ \ \ \Leftrightarrow\ \ \ x^2-8=0\\\\x^2-(2 √(2) )^2=0\ \ \ \Leftrightarrow\ \ \ (x-2 √(2) )(x+2 √(2) )=0\\\\x-2 √(2) =0\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x+2 √(2)=0\\\\x=2 √(2)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-2 \sqrt{2`

Answer 2

`y= (1)/(2)x^2-4\\ \\ y =0 \\ \\(1)/(2)x^2-4 =0 \ \ / \cdot 2\\ \\x^2-8=0 \\ \\(x-√(8))(x+√(8))=0 \\ \\ x-√(8)= \ \ or \ \ x+√(8) = 0 \\ \\x=√(8) \ \ or \ \ x=-√(8) \\ \\x=√(4\cdot 2) \ \ or \ \ x= -√(4\cdot 2)\\ \\ x=2√(2) \ \ or \ \ x=-2√(2)`